Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 100011;
int n, m, q, c, dp[N][20], dis[N];
vector<int> a[N][20], g[N], now;
void dfs(int v, int par) {
  dis[v] = dis[par] + 1;
  dp[v][0] = par;
  sort(a[v][0].begin(), a[v][0].end());
  a[v][0].erase(unique(a[v][0].begin(), a[v][0].end()), a[v][0].end());
  a[v][0].resize(min(10, (int)a[v][0].size()));
  for (int i = 1; i < 20; i++) {
    dp[v][i] = dp[dp[v][i - 1]][i - 1];
    a[v][i].insert(a[v][i].begin(), a[v][i - 1].begin(), a[v][i - 1].end());
    a[v][i].insert(a[v][i].begin(), a[dp[v][i - 1]][i - 1].begin(),
                   a[dp[v][i - 1]][i - 1].end());
    sort(a[v][i].begin(), a[v][i].end());
    a[v][i].erase(unique(a[v][i].begin(), a[v][i].end()), a[v][i].end());
    a[v][i].resize(min(10, (int)a[v][i].size()));
  }
  for (int i = 0; i < g[v].size(); i++) {
    if (g[v][i] != par) dfs(g[v][i], v);
  }
}
int LCA(int v, int u) {
  if (dis[v] < dis[u]) swap(u, v);
  while (dis[v] - dis[u]) {
    int k = log2(dis[v] - dis[u]);
    now.insert(now.begin(), a[v][k].begin(), a[v][k].end());
    v = dp[v][k];
  }
  if (v == u) return v;
  for (int i = 19; i >= 0; i--) {
    if (dp[v][i] != dp[u][i]) {
      now.insert(now.begin(), a[v][i].begin(), a[v][i].end());
      now.insert(now.begin(), a[u][i].begin(), a[u][i].end());
      v = dp[v][i];
      u = dp[u][i];
    }
  }
  now.insert(now.begin(), a[v][0].begin(), a[v][0].end());
  now.insert(now.begin(), a[u][0].begin(), a[u][0].end());
  v = dp[v][0];
  return v;
}
int main() {
  scanf("%d%d%d", &n, &m, &q);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[v].push_back(u);
    g[u].push_back(v);
  }
  for (int i = 1; i <= m; i++) {
    scanf("%d", &c);
    a[c][0].push_back(i);
  }
  dfs(1, 0);
  while (q--) {
    int u, v, f;
    scanf("%d%d%d", &v, &u, &f);
    int s = LCA(u, v);
    now.insert(now.begin(), a[s][0].begin(), a[s][0].end());
    sort(now.begin(), now.end());
    now.erase(unique(now.begin(), now.end()), now.end());
    printf("%d ", min((int)now.size(), f));
    for (int i = 0; i < f and i < now.size(); i++) printf("%d ", now[i]);
    printf("\n");
    now.clear();
  }
}